Question 833364: Please help me write an equation for this problem: In a triangle, one angle is twice the smallest angle and the third angle is 40 more than the smallest angle. Find all three angles.
Answer by Leaf W.(135)   (Show Source):
You can put this solution on YOUR website! Hello!
*
I will refer to the three angles as they are called in the problem -- "one angle," "the smallest angle," and "the third angle."
*
First, let us find expressions for each of the three angles. Since the "one angle" and "the third angle" are both described in relation to "the smallest angle," it will be simplest to use "the smallest angle" as the variable (although you could choose any of the three to be represented by the variable).
*
Therefore, let us say that "the smallest angle" = x.
*
Since "one angle is twice the smallest angle," we can say that "one angle" = 2x.
*
Since "the third angle is 40 more than the smallest angle," we can say that "the third angle" = x + 40.
*
As you may know, the three angles in a triangle will always add up to 180 degrees. Therefore, we know that the sum of all three angles is 180. In other words, "the smallest angle" + "one angle" + "the third angle" = 180.
*
Now we can just replace the names of the angles with the expressions we came up with above: (x) + (2x) + (x + 40) = 180
*
Therefore, your equation for this problem would be x + 2x + x + 40 = 180. To solve, just simplify the equation, solve for x, and then plug in your value for x into the expressions for each of the angles that you came up with above.
|
|
|
